Integrand size = 27, antiderivative size = 104 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\frac {32 c \sqrt {c+d x^3}}{3 d^3}+\frac {64 c^2 \sqrt {c+d x^3}}{3 d^3 \left (8 c-d x^3\right )}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac {352 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^3} \] Output:
32/3*c*(d*x^3+c)^(1/2)/d^3+64/3*c^2*(d*x^3+c)^(1/2)/d^3/(-d*x^3+8*c)+2/9*( d*x^3+c)^(3/2)/d^3-352/9*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^3
Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.76 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2 \left (\frac {\sqrt {c+d x^3} \left (-488 c^2+41 c d x^3+d^2 x^6\right )}{-8 c+d x^3}-176 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )\right )}{9 d^3} \] Input:
Integrate[(x^8*Sqrt[c + d*x^3])/(8*c - d*x^3)^2,x]
Output:
(2*((Sqrt[c + d*x^3]*(-488*c^2 + 41*c*d*x^3 + d^2*x^6))/(-8*c + d*x^3) - 1 76*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]))/(9*d^3)
Time = 0.37 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {948, 100, 27, 90, 60, 73, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^6 \sqrt {d x^3+c}}{\left (8 c-d x^3\right )^2}dx^3\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{3/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {\int \frac {c d \sqrt {d x^3+c} \left (9 d x^3+104 c\right )}{8 c-d x^3}dx^3}{9 c d^3}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{3/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {\int \frac {\sqrt {d x^3+c} \left (9 d x^3+104 c\right )}{8 c-d x^3}dx^3}{9 d^2}\right )\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{3/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {176 c \int \frac {\sqrt {d x^3+c}}{8 c-d x^3}dx^3-\frac {6 \left (c+d x^3\right )^{3/2}}{d}}{9 d^2}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{3/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {176 c \left (9 c \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {6 \left (c+d x^3\right )^{3/2}}{d}}{9 d^2}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{3/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {176 c \left (\frac {18 c \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {6 \left (c+d x^3\right )^{3/2}}{d}}{9 d^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (\frac {64 c \left (c+d x^3\right )^{3/2}}{9 d^3 \left (8 c-d x^3\right )}-\frac {176 c \left (\frac {6 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {6 \left (c+d x^3\right )^{3/2}}{d}}{9 d^2}\right )\) |
Input:
Int[(x^8*Sqrt[c + d*x^3])/(8*c - d*x^3)^2,x]
Output:
((64*c*(c + d*x^3)^(3/2))/(9*d^3*(8*c - d*x^3)) - ((-6*(c + d*x^3)^(3/2))/ d + 176*c*((-2*Sqrt[c + d*x^3])/d + (6*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3* Sqrt[c])])/d))/(9*d^2))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 1.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.76
| method | result | size |
| pseudoelliptic | \(\frac {\frac {2 \left (d \,x^{3}+c \right )^{\frac {3}{2}}}{9}+\frac {32 c \sqrt {d \,x^{3}+c}}{3}+\frac {32 c^{2} \left (\frac {2 \sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}-\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 \sqrt {c}}\right )}{3}}{d^{3}}\) | \(79\) |
| default | \(\frac {2 \left (d \,x^{3}+c \right )^{\frac {3}{2}}}{9 d^{3}}-\frac {16 c \left (-2 \sqrt {d \,x^{3}+c}+6 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )\right )}{3 d^{3}}+\frac {64 c^{2} \left (\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 \sqrt {c}}\right )}{3 d^{3}}\) | \(105\) |
| risch | \(\frac {2 \left (d \,x^{3}+49 c \right ) \sqrt {d \,x^{3}+c}}{9 d^{3}}+\frac {16 c^{2} \left (-\frac {26 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 \sqrt {c}\, d}+\frac {4 c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{3 d}\right )}{d^{2}}\) | \(109\) |
| elliptic | \(\frac {64 c^{2} \sqrt {d \,x^{3}+c}}{3 d^{3} \left (-d \,x^{3}+8 c \right )}+\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d^{2}}+\frac {98 c \sqrt {d \,x^{3}+c}}{9 d^{3}}+\frac {176 i c \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}-8 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{18 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{27 d^{5}}\) | \(473\) |
Input:
int(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
Output:
2/3*(1/3*(d*x^3+c)^(3/2)+16*c*(d*x^3+c)^(1/2)+16*c^2*(2*(d*x^3+c)^(1/2)/(- d*x^3+8*c)-11/3*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)))/d^3
Time = 0.08 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.81 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\left [\frac {2 \, {\left (88 \, {\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + {\left (d^{2} x^{6} + 41 \, c d x^{3} - 488 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{9 \, {\left (d^{4} x^{3} - 8 \, c d^{3}\right )}}, \frac {2 \, {\left (176 \, {\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + {\left (d^{2} x^{6} + 41 \, c d x^{3} - 488 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{9 \, {\left (d^{4} x^{3} - 8 \, c d^{3}\right )}}\right ] \] Input:
integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x, algorithm="fricas")
Output:
[2/9*(88*(c*d*x^3 - 8*c^2)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + (d^2*x^6 + 41*c*d*x^3 - 488*c^2)*sqrt(d*x^3 + c)) /(d^4*x^3 - 8*c*d^3), 2/9*(176*(c*d*x^3 - 8*c^2)*sqrt(-c)*arctan(3*sqrt(-c )/sqrt(d*x^3 + c)) + (d^2*x^6 + 41*c*d*x^3 - 488*c^2)*sqrt(d*x^3 + c))/(d^ 4*x^3 - 8*c*d^3)]
\[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\int \frac {x^{8} \sqrt {c + d x^{3}}}{\left (- 8 c + d x^{3}\right )^{2}}\, dx \] Input:
integrate(x**8*(d*x**3+c)**(1/2)/(-d*x**3+8*c)**2,x)
Output:
Integral(x**8*sqrt(c + d*x**3)/(-8*c + d*x**3)**2, x)
Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2 \, {\left (88 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 48 \, \sqrt {d x^{3} + c} c - \frac {96 \, \sqrt {d x^{3} + c} c^{2}}{d x^{3} - 8 \, c}\right )}}{9 \, d^{3}} \] Input:
integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")
Output:
2/9*(88*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqr t(c))) + (d*x^3 + c)^(3/2) + 48*sqrt(d*x^3 + c)*c - 96*sqrt(d*x^3 + c)*c^2 /(d*x^3 - 8*c))/d^3
Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\frac {352 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{9 \, \sqrt {-c} d^{3}} - \frac {64 \, \sqrt {d x^{3} + c} c^{2}}{3 \, {\left (d x^{3} - 8 \, c\right )} d^{3}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{6} + 48 \, \sqrt {d x^{3} + c} c d^{6}\right )}}{9 \, d^{9}} \] Input:
integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x, algorithm="giac")
Output:
352/9*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 64/3*sqrt( d*x^3 + c)*c^2/((d*x^3 - 8*c)*d^3) + 2/9*((d*x^3 + c)^(3/2)*d^6 + 48*sqrt( d*x^3 + c)*c*d^6)/d^9
Time = 2.43 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\frac {98\,c\,\sqrt {d\,x^3+c}}{9\,d^3}+\frac {176\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{9\,d^3}+\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^2}+\frac {64\,c^2\,\sqrt {d\,x^3+c}}{3\,d^3\,\left (8\,c-d\,x^3\right )} \] Input:
int((x^8*(c + d*x^3)^(1/2))/(8*c - d*x^3)^2,x)
Output:
(98*c*(c + d*x^3)^(1/2))/(9*d^3) + (176*c^(3/2)*log((10*c + d*x^3 - 6*c^(1 /2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/(9*d^3) + (2*x^3*(c + d*x^3)^(1/2)) /(9*d^2) + (64*c^2*(c + d*x^3)^(1/2))/(3*d^3*(8*c - d*x^3))
\[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\frac {\frac {656 \sqrt {d \,x^{3}+c}\, c^{2}}{45}-\frac {82 \sqrt {d \,x^{3}+c}\, c d \,x^{3}}{9}-\frac {2 \sqrt {d \,x^{3}+c}\, d^{2} x^{6}}{9}+\frac {12672 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{3} d^{2}}{5}-\frac {1584 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{2} d^{3} x^{3}}{5}}{d^{3} \left (-d \,x^{3}+8 c \right )} \] Input:
int(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x)
Output:
(2*(328*sqrt(c + d*x**3)*c**2 - 205*sqrt(c + d*x**3)*c*d*x**3 - 5*sqrt(c + d*x**3)*d**2*x**6 + 57024*int((sqrt(c + d*x**3)*x**5)/(64*c**3 + 48*c**2* d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c**3*d**2 - 7128*int((sqrt(c + d*x **3)*x**5)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c**2 *d**3*x**3))/(45*d**3*(8*c - d*x**3))